Adjustable voltage regulator (TC1174 device) questions

Log in or register to post comments
Go To Last Post
7 posts / 0 new
Author
Message
unebonnevie
unebonnevie's picture
Level: Resident
Joined: Thu. Oct 30, 2008
Posts: 748 View posts
#1
Posted by unebonnevie: Mon. Mar 14, 2011 - 01:17 AM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hi,

I am planning to use the TC1174 to regulate a 5V down to 3.6V. Please see the attached schematic.

Vin = 5V and max. 500mA can ne drawn from this Vin.
Vout = 3.6V and max 300mA (TC1174's max current output) can be drawn from Vout.

The max power (worst case) out is:

Pout = VI = 3.6V * 300mA = 3.6 * 0.3 = 1.08W

Note that I will adjust R2 accordingly to get a Vout = 3.6V.

My question is: Is 1/2W resistor rating for R1 and R2 enough, since there is current divider going through the Vout and the resistors?

Thanks

Attachment(s): 

Image icon tc1174_schematic.jpg

Tags:

Learning and Information, General Electronics
Top
Kartman
Kartman's picture
Level: 10k+ Postman
Joined: Thu. Dec 30, 2004
Posts: 24785 View posts
Location: Melbourne,Australia
Posted by Kartman: Mon. Mar 14, 2011 - 03:43 AM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

If R1and 2 are 470k, then what current is flowing through them? Then calculate the wattage W= V*I. Have a read of ohms law and you'll not have to ask this question again.

Short answer- half watt will be more than suffient.

Top
unebonnevie
unebonnevie's picture
Level: Resident
Joined: Thu. Oct 30, 2008
Posts: 748 View posts
Posted by unebonnevie: Mon. Mar 14, 2011 - 04:27 AM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Kartman wrote:
If R1and 2 are 470k, then what current is flowing through them? Then calculate the wattage W= V*I. Have a read of ohms law and you'll not have to ask this question again.

Short answer- half watt will be more than suffient.

I think you simplify it too much there. At least two KCL loops, R1 with the cap, and one KCL loop on R2, thus, my asking.

Top
Kartman
Kartman's picture
Level: 10k+ Postman
Joined: Thu. Dec 30, 2004
Posts: 24785 View posts
Location: Melbourne,Australia
Posted by Kartman: Mon. Mar 14, 2011 - 08:14 AM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

On the contrary, i think you're not considering the direction of the currents and the voltages involved. How much voltage is required across a 470k resistor in order to disipate 0.5 W?

Top
Chris-Mouse
Chris-Mouse's picture
Level: Hangaround
Joined: Tue. Apr 20, 2010
Posts: 112 View posts
Location: Toronto, Canada
Posted by Chris-Mouse: Mon. Mar 14, 2011 - 05:42 PM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Vout is essentially a DC value. Once C1 has charged, it plays no further part in power dissipation in the circuit.
The adj terminal is a high impedance input. The current into and out of it is also essentially zero.
R1 and R2 form a voltage divider from Vout to ground, with essentially no other path for the current to flow. Calculating voltages, currents, and power dissipation is a simple Ohms law and power law problem.

Top
ka7ehk
ka7ehk's picture
Level: 10k+ Postman
Joined: Fri. Nov 22, 2002
Posts: 21821 View posts
Location: Tangent, OR, USA
Posted by ka7ehk: Mon. Mar 14, 2011 - 06:09 PM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Consider, as others have suggested, how much voltage across a 470K resistor is needed to dissipate 0.5W. Then, ask yourself what the largest voltage is in your circuit? Is there any way to exceed this voltage even if it is turned on with all capacitors discharged?

I think you will find that 0.5W resistors are really quite excessive. I think that you will find that even 100mW resistors are more than sufficient.

Also, not quite sure of where you get the idea of "current divider" in the output. In the sense that the term "current divider" is normally defined, I don't see any.

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

Top
unebonnevie
unebonnevie's picture
Level: Resident
Joined: Thu. Oct 30, 2008
Posts: 748 View posts
Posted by unebonnevie: Tue. Mar 15, 2011 - 11:33 PM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Thanks, all for tips and analysis that I don't really have to consider the cap, and ultimately, it's just a resistive voltage divider.

Top